Uniform sums and Euler’s number

Suppose I sample values uniformly at random from the unit interval.
How many samples should I expect to take before the sum of my sampled values exceeds unity?1

Let \(N\) be the (random) number of samples taken when the sum first exceeds unity.
Then \(N\) has expected value \(E[N]\) equal to Euler’s number\(e\approx2.718282\).
This can be verified approximately via simulation:

To see why \(E[N]=e\), let \((X_i)_{i=1}^\infty\) be an infinite sequence of random variables with uniform distributions over the unit interval.
Then the probability that \(N\) exceeds any non-negative integer \(n\) is
$$\Pr(N>n)=\Pr(X_1+X_2+\cdots+X_n<1).$$
Consider the unit (hyper)cube in \(\mathbb{R}^n\).
Its vertices comprise the origin, the standard basis vectors \(e_1,e_2,\ldots,e_n\), and the sums of two or more of these basis vectors.
The convex hull of \(\{0,e_1,e_2,\ldots,e_n\}\) forms an \(n\)-simplex with volume \(1/n!\).
The interior of this simplex is precisely the set
$$\{X_1,X_2,\ldots,X_n\in[0,1]:X_1+X_2+\cdots+X_n<1\}.$$
It follows that \(\Pr(X_1+X_2+\cdots+X_n<1)=1/n!\) and therefore \(\Pr(N>n)=1/n!\) from above.
Now
$$\Pr(N=n)=\Pr(N>n-1)-\Pr(N>n)$$
for each \(n\ge1\).
Thus, since \(\Pr(N>0)=1\) (and, by convention, \(0!=1\)), we have
$$\begin{align} E[N] &= \sum_{n=1}^\infty n\Pr(N=n) \\ &= \sum_{n=1}^\infty n\left(\Pr(N>n-1)-\Pr(N>n)\right) \\ &= \Pr(N>0)+\sum_{n=1}^\infty\Pr(N>n) \\ &= 1+\sum_{n=1}^\infty\frac{1}{n!} \\ &= \sum_{n=0}^\infty\frac{1}{n!} \\ &= e. \end{align}$$
The final equality comes from evaluating the Maclaurin series expansion of \(e^x\) at \(x=1\).